# Calculating the number of energy states in a particular momentum direction

Given is a 2D scenario of a particle within a container:

The circumference shows all the possible 2D spatial directions for a given momentum value of the particle. One of those momentum directions is $$p_e$$.

From what I understand, the number of states within that $$p_e$$ is given by: $$n_e = \sqrt{\frac{L_x \cdot p_{e,x}}{h}^2 + \frac{L_y \cdot p_{e,y,}}{h}^2}$$ Where $$p_{e,x}$$ and $$p_{e,y}$$ are the projections of $$p_e$$ in the $$x$$ and $$y$$ coordinates.

Since $$p_e$$ has a spatial direction, this means that the container has a length in that same spatial direction. Let's call that container length in that same spatial direction $$L_e$$, illustrated as follows:

Does this $$L_e$$ have any relationship with the number of states in momentum $$p_e$$? For example, is the following equation valid? $$n_e = \sqrt{\frac{L_x \cdot p_{e,x}}{h}^2 + \frac{L_y \cdot p_{e,y,}}{h}^2} = \frac{L_e p_e}{h}$$

EDIT: I forgot to state to use periodic boundary condition in such a case, since rigid boundaries gives zero momentum to particles. So my question is: is it possible to calculate the number of states in a single momentum that has just one direction in p-space? If so, how is this derived from the formula: $$n=\frac{V \cdot p^3}{h^3}$$

• Don't all eigenstates of a particle in a box have zero momentum? – Ivan Neretin Nov 6 '18 at 15:02
• Apologies, I should have stated to use periodic boundary conditions – JohnnyGui Nov 14 '18 at 21:56